$h(t) = -5t^{2}+7t$ $f(x) = 6x^{3}-2x^{2}+2(h(x))$ $ h(f(0)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = 6(0^{3})-2(0^{2})+2(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = -5(0^{2})+(7)(0)$ $h(0) = 0$